Question

Calculate the energy released when three \( \alpha \)-particles combined to form a \( \text{C}^{12} \) nucleus, the mass defect is

• A. \( 0.0063 \, \text{u} \)
• B. \( 0.0062 \, \text{u} \)
• C. \( 0.0021 \, \text{u} \)
• D. \( 0.5 \, \text{u} \)

Hint:

To calculate the energy released during a nuclear reaction, use \( E = \Delta m c^2 \), where \( \Delta m \) is the mass defect.

Answer 1

The Correct Option is B

The energy released during the formation of a nucleus can be calculated using the equation \( E = \Delta m c^2 \), where \( \Delta m \) is the mass defect. Given the mass defect, we calculate the energy released.

Step 2: Conclusion.
The energy released is calculated to be \( 0.0062 \, \text{u} \), corresponding to option (b).